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What is the equation, in point-slope form, of the line that passes through the point (3,-4) and is perpendicular to the line represented by y=2/5x+6?

a) y = -5x - 19
b) y = -5x + 6
c) y = -5x - 2
d) y = 5x - 6

User Dgimenes
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1 Answer

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Final answer:

The equation, in point-slope form, of the line that passes through the point (3, -4) and is perpendicular to the line represented by y = (2/5)x + 6 is y = -5/2x + 7/2.

Step-by-step explanation:

To find the equation of a line that is perpendicular to another line, we need to know the negative reciprocal of the slope of the given line. Given that the equation of the given line is y = (2/5)x + 6, the slope of this line is 2/5. The negative reciprocal of 2/5 is -5/2. So, the slope of the perpendicular line is -5/2.

Now, we can use the point-slope form of a linear equation y-y1 = m(x-x1), where (x1, y1) is a point on the line, and m is the slope of the line. Using the point (3, -4) and the slope -5/2, we get the equation y - (-4) = -5/2(x - 3), which simplifies to y + 4 = -5/2x + 15/2. Solving this equation for y gives us y = -5/2x + 15/2 - 8/2, which simplifies to y = -5/2x + 7/2. Therefore, the equation, in point-slope form, of the line that passes through the point (3, -4) and is perpendicular to the line represented by y = (2/5)x + 6 is y = -5/2x + 7/2.

User Blackbam
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